Question

Find the number of proton donor acid(s) from the following:
\[B{{(OH)}_{3}}\], $Mg{{(OH)}_{2}}$, $Si{{(OH)}_{4}}$, $S{{O}_{2}}{{(OH)}_{2}}$, \[Ba{{(OH)}_{2}}\]

Answer 1

Hint: According to the Bronsted-Lowry concept of acids and bases, any species donating proton (${{H}^{+}}$ ion) to other species are acids whereas Lewis concept defines acids as the species which can accept a pair of electrons from other molecules.


Complete answer:
Proton donor acids can donate one or more ${{H}^{+}}$ ions to other molecules (bases). When proton donor acids are dissolved in water, they dissociate to give ${{H}^{+}}$ ions. Hydration of protons by water molecules then forms hydronium ions.
\[\begin{align}
& HA\to {{H}^{+}}+{{A}^{-}} \\
& {{H}^{+}}+{{H}_{2}}O\to {{H}_{3}}{{O}^{+}} \\
\end{align}\]
Now let us examine all the compounds given one by one to find out the total number of proton donor acids.

- \[B{{(OH)}_{3}}\]: Orthoboric acid or boric acid. It is a monobasic Lewis acid. it does release a proton but rather accepts a pair of electrons. It belongs to the class of aprotic acids.
- $Mg{{(OH)}_{2}}$: Magnesium hydroxide is not an acid but a base. It is a weaker base than alkali metal bases like NaOH, KOH, etc. $Mg{{(OH)}_{2}}$ is a base according to the Bronsted-Lowry concept of bases. When $Mg{{(OH)}_{2}}$ is dissolved in water, it dissociates into $M{{g}^{2+}}$and $O{{H}^{-}}$ ions.
$Mg{{(OH)}_{2}}\to M{{g}^{2+}}(aq)+2O{{H}^{-}}$
- $Si{{(OH)}_{4}}$: Silicic acid or orthosilicic acid is a weak acid. Although it is a weak acid it gives a proton and hence, is a proton donor acid.
$Si{{(OH)}_{4}}+{{H}_{2}}O\to Si{{(OH)}_{3}}{{O}^{-}}+{{H}^{+}}$
- $S{{O}_{2}}{{(OH)}_{2}}$: It is sulphuric acid i.e., ${{H}_{2}}S{{O}_{4}}$. Sulphuric acid is a very strong acid. It ionizes first into ${{H}^{+}}$and $HSO_{4}^{-}$. Then, ${{H}^{+}}$ ions are surrounded by water molecules to form hydronium ions.
$\begin{align}
& {{H}_{2}}S{{O}_{4}}\to {{H}^{+}}+HSO_{4}^{-} \\
& {{H}^{+}}+{{H}_{2}}O\to {{H}_{3}}{{O}^{+}} \\
\end{align}$
$HSO_{4}^{-}$ further dissociates to give ${{H}^{+}}$ and $SO_{4}^{2-}$ ions. The overall reaction of sulphuric acid in water is given as:
\[{{H}_{2}}S{{O}_{4}}+2{{H}_{2}}O\to 2{{H}_{3}}{{O}^{+}}+SO_{4}^{2-}\]
- \[Ba{{(OH)}_{2}}\] : Barium hydroxide solution is a strong base. It completely dissociates in water into $B{{a}^{2+}}$ and \[O{{H}^{-}}\] ions. Hence, it is a Bronsted-Lowry base.
\[Ba{{(OH)}_{2}}\to B{{a}^{2+}}+2O{{H}^{-}}\]

The proton donor acids are sulphuric acid ($S{{O}_{2}}{{(OH)}_{2}}$) and silicic acid ($Si{{(OH)}_{4}}$).

Therefore, the number of proton donor acids is two.

Note:
Note that \[B{{(OH)}_{3}}\] is also an acid but it is a Lewis acid. It is electron deficient in nature as boron has only six electrons in its outermost shell. Therefore, when dissolved in water, it accepts a pair of electrons from water in the form of \[O{{H}^{-}}\] ion.